$\text{The sum of the spin only magnetic moment values (in B.M.) of}\left[\mathrm{Mn}\right(\mathrm{Br}\left){\mathrm{₆}}_{}\right]{}^{3-}\text{and}\left[\mathrm{Mn}\right(\mathrm{CN}\left){\mathrm{₆}}_{}\right]{}^{3-}\text{is \_\_\_\_\_\_\_\_.}$

The correct answer is 7.72.

To find the sum of the spin-only magnetic moment values of the two coordination complexes, $\left[\mathrm{Mn}\right(\mathrm{Br}{)}_6{]}^{3-}$ and $\left[\mathrm{Mn}\right(\mathrm{CN}{)}_6{]}^{3-}$, we need to determine the oxidation state of manganese in both complexes, their d-electron configurations, and the number of unpaired electrons under the influence of weak-field and strong-field ligands.

Step 1: Determine the oxidation state of Mn in both complexes
For $\left[\mathrm{Mn}\right(\mathrm{Br}{)}_6{]}^{3-}$:
Let the oxidation state of Mn be x. Since bromide (${\mathrm{Br}}^{-}$) is a monodentate anionic ligand with a charge of -1:
$x+6(-1)=-3\Rightarrow x=+3$
For $\left[\mathrm{Mn}\right(\mathrm{CN}{)}_6{]}^{3-}$:
Let the oxidation state of Mn be y. Since cyanide (${\mathrm{CN}}^{-}$) is also a monodentate anionic ligand with a charge of -1:
$y+6(-1)=-3\Rightarrow y=+3$
In both complexes, manganese is in the +3 oxidation state (${\mathrm{Mn}}^{3+}$).

Step 2: Electronic configuration of ${\mathrm{Mn}}^{3+}$
The atomic number of manganese (Mn) is 25.
The electronic configuration of neutral Mn is $\left[\mathrm{Ar}\right]3d^{5}4s^2$.
Thus, the electronic configuration of ${\mathrm{Mn}}^{3+}$ is $\left[\mathrm{Ar}\right]3d^4$, containing 4 d-electrons.

Step 3: Analyze crystal field splitting and determine unpaired electrons (n)
Case A: $\left[\mathrm{Mn}\right(\mathrm{Br}{)}_6{]}^{3-}$
Bromide (${\mathrm{Br}}^{-}$) is a weak-field ligand. According to Crystal Field Theory, the crystal field splitting energy (${\Delta }_o$) is less than the pairing energy (P). Consequently, pairing of electrons does not occur, resulting in a high-spin complex.
The electronic configuration in the octahedral crystal field is:
$t_{2g}^3{e}_g^1$
Number of unpaired electrons ($n_1$) = 4.
The spin-only magnetic moment (${\mu }_1$) is:
${\mu }_1=\sqrt{n_1(n_1+2)}=\sqrt{4(4+2)}=\sqrt{24}\approx 4.899\text{B.M.}$

Case B: $\left[\mathrm{Mn}\right(\mathrm{CN}{)}_6{]}^{3-}$
Cyanide (${\mathrm{CN}}^{-}$) is a strong-field ligand. The crystal field splitting energy (${\Delta }_o$) is greater than the pairing energy (P). This causes the electrons to pair up in the lower energy $t_{2g}$ orbitals before entering the $e_g$ orbitals, resulting in a low-spin complex.
The electronic configuration in the octahedral crystal field is:
$t_{2g}^4{e}_g^0$
Here, out of 4 d-electrons, two are paired in one $t_{2g}$ orbital, leaving two unpaired electrons.
Number of unpaired electrons ($n_2$) = 2.
The spin-only magnetic moment (${\mu }_2$) is:
${\mu }_2=\sqrt{n_2(n_2+2)}=\sqrt{2(2+2)}=\sqrt{8}\approx 2.828\text{B.M.}$

Step 4: Calculate the sum of the magnetic moments
$\text{Total spin-only magnetic moment}={\mu }_1+{\mu }_2$
$=4.899+2.828=7.727\approx 7.72\text{B.M.}$